Spring-loaded switch

✔ 最佳答案

讓我嘗試回答你的問題(我也不知到concept是否完全正確)
首先,我估你是說LC circuit.
個(self-)induced emf 在open (separate) switch時較close switch 大很多.
因為在close switch時, current increase slowly( I= [1-exp(-Rt/L)] E/R ) E= voltage supply-->induced emf 不太大.( induced emf,є = -L dI/dt)
但在open switch時, current suddently be cut off--> є become very large
Therefore, "є" is very large when opening switch.
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接著,說說potential difference(pd) btw the switch.
Before the switch is closed, pd btw the switch= voltage supply(E), because resistance(R) btw switch is extremely large and draw all the potential.
While the switch just start opening, pd btw the switch=E+є which is the 1st main point of this problem.
Therefore, "pd btw the switch" is larger when opening switch than that of closing switch.
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pd btw the switch 太大會導致1個問題, 就係個開關有火花(spark or electric arc)( electric field= pd/distance) (electric breakdown in air = 3x10^6 V/m)
這個火花當然要distance btw the contacts of switch 來配合.
When the switch is opened slowly, a short distance btw the contacts of switch will remain a longer time than that of opening quickly. (The 2nd main point)
Therefore, when the contacts of the switch is separated quickly, the spark can be avoided.
到這裡不知有冇有解答了你的問題
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據我所知,還有2個方法避免火花.
1. connecting a capacitor in parallel with the switch
2. connecting a a neon lamp in parallel with the inductor(個neon lamp 要大過50V OR 大過100V先開到, 所open switch 先會著燈)
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有火花有咩問題:
1.危險,電器有火花當然危險啦
2.開關鍵入面可能有d野會因為溫度高而熔解
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conclusion
"є" is very large when opening switch.
"pd btw the switch" is larger when opening switch.
The spark can be prevented when opening switch quickly by using Spring-loaded switch.
The book make a mistake in induced emf.
I believe that opening switch quickly can almost prevent electrons move btw the contacts of the switch so the switch arms will not be melted.
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sorry about my bad English
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additional info
http://www.school-for-champions.com/science/static_sparks.htm

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